The tee for the sixth hole on a golf course is 400 yards from the tee. On that hole, Marsha hooked her ball to the left, as sketched below. Find the distance between Marsha’s ball and the hole to the nearest tenth of a yard. Answer any time! :D

  181.8 yd

Step-by-step explanation:

The law of cosines is good for this. It tells you for triangle sides 'a' and 'b' and included angle C, the length of 'c' is given by ...

  c^2 = a^2 +b^2 -2ab·cos(C)

For the given geometry, this is ...

  c^2 = 400^2 +240^2 -2(400)(240)cos(16°) ≈ 33,037.75

  c ≈ √33037.75 ≈ 181.8 . . . yards

Marsha's ball is about 181.8 yards from the hole.

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation: